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Grade: 12
        

what is the diff. of      sin6x+cos6x/sin3x+cos3x

10 years ago

Answers : (3)

ashish kumar
17 Points
							

 

Dear Mustafa,

Its diferentiation will be : -

( 6cos6x  - 6sin6x ) /(3cos3x - 3sin3x )

or, 2 ( cos6x - sin6x ) / ( cos3x - cos3x )

Thank you.

10 years ago
ashish kumar
17 Points
							


Dear Mustafa,

Its diferentiation will be : -

( 6cos6x  - 6sin6x ) /(3cos3x - 3sin3x )

or, 2 ( cos6x - sin6x ) / ( cos3x - cos3x )

Thank you.

10 years ago
Badiuddin askIITians.ismu Expert
147 Points
							

Hi mustafa

y=(sin6x+cos6x)/(sin3x+cos3x)

dy/dx ={(sin3x+cos3x)(6cos6x-6sin6x)-(3cos3x-3sin3x)(sin6x+cos6x)}/(sin3x+cos3x)

                 

Now simplyfy

dy/dx =(3 cos9x -9sin3x)/ (sin3x+cos3x)2

 


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Askiitians Experts

Badiuddin

10 years ago
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